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A coil has inductance of $0.4 \mathrm{H}$ and resistance of $8 \Omega$. It is connected to an $\mathrm{AC}$ source with peak emf $4 \mathrm{~V}$ and frequency $\frac{30}{\pi} \mathrm{Hz}$. The average power dissipated in the circuit is
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Verified Answer
The correct answer is:
$0.1 \mathrm{~W}$
Average power dissipated is
$$
\begin{aligned}
& P_{\text {avg }}=V_{\text {rms }} I_{\mathrm{rms}} \cos \phi \\
& =\frac{V_{\max }}{\sqrt{2}} \times\left(\frac{V_{\max }}{\sqrt{2}}\right) \times \frac{1}{Z} \times \frac{R}{Z} \\
& =\frac{V_{\max }^2}{2} \times \frac{R}{Z^2}=\frac{V_{\max }^2}{2} \times \frac{R}{\left(\sqrt{X_L^2+R^2}\right)} \\
& =\frac{(4)^2}{2} \times \frac{8}{\left.\sqrt{(0.4 \times 60)^2+8^2}\right)^2} \\
& =\frac{16 \times 8}{2 \times\left(24^2+8^2\right)}=\frac{64}{640}=\frac{1}{10}=0.1 \mathrm{~W}
\end{aligned}
$$
$$
\begin{aligned}
& P_{\text {avg }}=V_{\text {rms }} I_{\mathrm{rms}} \cos \phi \\
& =\frac{V_{\max }}{\sqrt{2}} \times\left(\frac{V_{\max }}{\sqrt{2}}\right) \times \frac{1}{Z} \times \frac{R}{Z} \\
& =\frac{V_{\max }^2}{2} \times \frac{R}{Z^2}=\frac{V_{\max }^2}{2} \times \frac{R}{\left(\sqrt{X_L^2+R^2}\right)} \\
& =\frac{(4)^2}{2} \times \frac{8}{\left.\sqrt{(0.4 \times 60)^2+8^2}\right)^2} \\
& =\frac{16 \times 8}{2 \times\left(24^2+8^2\right)}=\frac{64}{640}=\frac{1}{10}=0.1 \mathrm{~W}
\end{aligned}
$$
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